Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

Scientists want to place a 2800.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2695.0 m/s in a perfectly circular orbit. Here is some information that may help solve this problem: mass of mars = 6.4191 x 10^23 kg radius of mars = 3.397 x 10^6 m G = 6.67428 x 10^-11 N-m^2/kg^2 What should the speed of the orbit be, if we want the satellite to take 8 times longer to complete one full revolution of its orbit?

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
My radius is 5.89876*10^6 m. I'm aware that T = (2pi/sqrt(GM))*r^(3/2) and that circumference = 2pi*r However, when try to solve the problem, I keep coming up with the wrong answer.
@Algebraic! Would you mind helping on this last one?
I solved for T, and then multiplied that time by 8. I then found the circumference of the orbit. I divided circumference/8T

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I think what you need to do is compare 2*pi*r/V to 16 *pi*r/V
you can use the result from the other problem we did to write r in terms of V I believe...
I'll be back to check and see what you have to say. :-) Thank you again
\[T = \frac{2 \pi r }{ V} = 2 \pi G\frac{ M _{mars} }{ V^3 }\]
so the period goes like the inverse cube of velocity....
\[\frac{ 8T }{ T } = \frac{ 2695^3 }{ V^3}\] \[2695^3 \frac{1}{8} = V^3 \] aka half the velocity...
I was thinking now that if the satellite changes velocity, it falls out of orbit and into a different one. If it takes 8 times as long, does the radius (as well as the orbit) change?
I'm still working through it...
yes.
It's really more like... you increase the orbital radius, force from gravity is weaker, centripetal force is less, velocity is less... so distance (circumference) is greater and velocity is smaller... that's what that expression I used shows... (I hope)

Not the answer you are looking for?

Search for more explanations.

Ask your own question